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Thread: solving equation

  1. #1
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    solving equation

    am i using a valid alternative solution?

    question:

    $\displaystyle \sin x+\cos\frac{x}{2}=0$

    correct solution:

    $\displaystyle \sin(2\cdot\frac{x}{2})+cos\frac{x}{2}=0$

    $\displaystyle \2sin\frac{x}{2}\cos\frac{x}{2}+cos\frac{x}{2}=0$

    $\displaystyle \cos\frac{x}{2}(\2sin\frac{x}{2}+1)=0$

    $\displaystyle \cos\frac{x}{2}=0, \sin\frac{x}{2}=-\frac{1}{2}$

    since $\displaystyle 0\leq x<2\pi, \sin\frac{x}{2}\geq 0$

    $\displaystyle \sin\frac{x}{2}\neq -\frac{1}{2}$

    $\displaystyle x=\pi$

    my solution (is this correct?):

    $\displaystyle \sin x+\sin(\frac{\pi}{2}\cdot\frac{x}{2})=0$

    $\displaystyle 2\sin\frac{x+\pi}{4}=0, \cos\frac{3x-\pi}{4}=0$

    $\displaystyle \frac{\pi}{4}\leq\frac{x+\pi}{4}<\frac{3}{4}\pi, /frac{\pi}{4}\leq\frac{3x-\pi}{4}<\frac{5}{4}\pi$

    $\displaystyle \frac{x+\pi}{4}=0, \frac{3x-\pi}{4}=\frac{\pi}{2}$

    since $\displaystyle 0\leq x<2\pi, x\neq -\pi$

    $\displaystyle x=\pi$
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: solving equation

    Quote Originally Posted by muddywaters View Post
    my solution (is this correct?):
    $\displaystyle \sin x+\sin(\frac{\pi}{2}\cdot\frac{x}{2})=0$
    What have you done here?

    An altenative solution could be to use the fact that $\displaystyle \cos\left(\frac{x}{2}\right)=\sin\left(\frac{\pi}{ 2}-\frac{x}{2}\right)$
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  3. #3
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    Re: solving equation

    Quote Originally Posted by Siron View Post
    What have you done here?

    An altenative solution could be to use the fact that $\displaystyle \cos\left(\frac{x}{2}\right)=\sin\left(\frac{\pi}{ 2}-\frac{x}{2}\right)$
    i'm sorry you're right. typo >< it should be $\displaystyle \cos\left(\frac{x}{2}\right)=\sin\left(\frac{\pi}{ 2}-\frac{x}{2}\right)$
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  4. #4
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    Re: solving equation

    with that, is it a valid alternative solutioN?
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  5. #5
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    Re: solving equation

    ish. just realized the mistakes in the first post. it should me liek this:

    my solution (is this correct?):

    $\displaystyle \sin x+\sin(\frac{\pi}{2}-\frac{x}{2})=0$

    $\displaystyle 2\sin\frac{x+\pi}{4}\cos\frac{3x-\pi}{4}=0$

    $\displaystyle 2\sin\frac{x+\pi}{4}=0, \cos\frac{3x-\pi}{4}=0$

    $\displaystyle \frac{\pi}{4}\leq\frac{x+\pi}{4}<\frac{3}{4}\pi, \frac{\pi}{4}\leq\frac{3x-\pi}{4}<\frac{5}{4}\pi$

    $\displaystyle \frac{x+\pi}{4}=0, \frac{3x-\pi}{4}=\frac{\pi}{2}$

    since $\displaystyle 0\leq x<2\pi, x\neq -\pi$

    $\displaystyle x=\pi$
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: solving equation

    Your solution looks fine to me.
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